a solid lies between planes perpendicular to x-axis at x=0 and x=17,cross section perpendicular to axis interval 17 greater or equal x & x greater or equal 0 are squares of diagonals go from parabolas y=-2x^(0.5) & y=2x^(0.5). find the volume of the solid

To find the volume of the solid between the given planes and with cross-sections perpendicular to the x-axis, we need to integrate the areas of the cross-sections over the interval from x=0 to x=17.

First, let's find the equation of the diagonal of each square cross-section. We are given that the diagonals go from parabolas y = -2x^(0.5) and y = 2x^(0.5).

The diagonal of a square can be found by using the Pythagorean theorem. Therefore, we have:

(diagonal)^2 = side^2 + side^2 = 2(side^2)

Let's find the diagonal for each cross-section at a given x-value:
For the cross-section at x = k, the side length of the square is y(k) = 2k^(0.5) - (-2k^(0.5)) = 4k^(0.5).

Therefore, the diagonal of each cross-section at x = k is:
(diagonal)^2 = 2(4k^(0.5))^2 = 32k

Now, we need to find the area of the cross-section at each x-value. The area of a square is equal to the side length squared, so the area of each cross-section at x = k is:
(area) = (side)^2 = (4k^(0.5))^2 = 16k

To find the volume of the solid, we integrate the areas of the cross-sections over the interval from x = 0 to x = 17:
(volume) = ∫[0,17] 16k dk

Integrating, we get:
(volume) = 16∫[0,17] k dk
(volume) = 16 * (1/2) * k^2 |[0,17]
(volume) = 8 * (17^2 - 0^2)
(volume) = 8 * 289
(volume) = 2312 cubic units

Therefore, the volume of the solid is 2312 cubic units.